I am really having a hard time in this intro to real analysis class. I feel as if I'm the only one in class who isn't getting it. I have an extremely hard time thinking abstractly and constructing my own proofs. I know I need a lot of practice. Here is the problem we have to prove:

Claim: Let A be a nonempty subset of R (all real numbers -- how do I type the symbol for real numbers?). If α = sup A is finite, show that for each ε > 0, there is an a in A such that α – ε < a ≤ α.

My attempt of a proof: Assume α = sup A is finite. Then A is bounded above because it is not empty and its supremum is finite (by the definition that if E is a nonempty subset of R (all reals), we set sup E = ∞ if E is not bounded above). [my question is where does the “ε” come from?] By definition of supremum, there is an element ß in R such that ß < α and ß is not an upper bound. In this case let ε be the ß where ε > 0. Knowing α is the supremum, ε < α, so there is an element a in A such that ε < a ≤ α or α – ε < a ≤ α.

*I also need to prove the converse of this statement which is:
"Let A be a nonempty subset of R (all real numbers) that is bounded above by α. Prove that if for every ε > 0 there is an a in A such that α – ε < a ≤ α, then α = sup A."

When proving the converse, isn't it just basically working backwards?
So I would write: Assume that for every ε > 0 there is an a in A such that α – ε < a ≤ α.
A is nonempty and bounded above by α (given). Then α = sup A is finite by the definition of supremum.

I feel really confused and lost here. I'm really afraid of this class. I need to pass it because it is only offered every 2 years.

Any help, suggestions, and guidance is greatly appreciated.
Thank you.

Sep 28th 2008, 03:47 PM

Plato

First. What does it mean to say that x is an upper bound for set A?
Second. What does it mean to say y is not an upper bound of set A?
Third. What does it mean to say z is the least upper bound (the sup) of set A?
Finally. If z=sup(A) is it possible that is an upper bound if ?

Sep 28th 2008, 03:51 PM

ThePerfectHacker

Quote:

Originally Posted by ilikedmath

Claim: Let A be a nonempty subset of R (all real numbers -- how do I type the symbol for real numbers?). If α = sup A is finite, show that for each ε > 0, there is an a in A such that α – ε < a ≤ α.

If is the supremum then cannot be the maximum since it is smaller than and is the supremum. Therefore there is so that .

Sep 28th 2008, 08:36 PM

ilikedmath

My work so far

Here is what I have so far:

Proof for 1st statement:

Let A be a nonempty subset of R. Assume alpha = sup A is finite. For any a in A, a < alpha by the definition of supremum. Since alpha is the supremum, alpha - epsilon cannot be an upper bound of A given any epsilon > 0. Since alpha - epsilon is not an upper bound, there exists an a in A such that
a > alpha - epsilon. Thus (alpha - epsilon) < a ≤ alpha. QED.

To prove the converse of the 1st statement should've been simply working backwards, right? I got stuck though:

Proof of converse:
Let A be a nonempty subset bounded above by alpha. Assume for every epsilon > 0 there is an a in A such that (alpha - epsilon) < a ≤ alpha. Clearly (alpha - epsilon) < alpha, and so (alpha - epsilon) is not an upper bound.

Now where can I go from there? (That is, if I am even on the right track(Surprised))